hdu 2227(树状数组+dp) Find the nondecreasing subsequences

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1844 Accepted Submission(s): 677

Problem Description

How many nondecreasing subsequences can you find in the sequence S = {s1, s2, s3, ...., sn} ? For example, we assume that S = {1, 2, 3}, and you can find seven nondecreasing subsequences, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}.

Input

The input consists of multiple test cases. Each case begins with a line containing a positive integer n that is the length of the sequence S, the next line contains n integers {s1, s2, s3, ...., sn}, 1 <= n <= 100000, 0 <= si <= 2^31.

Output

For each test case, output one line containing the number of nondecreasing subsequences you can find from the sequence S, the answer should % 1000000007.

Sample Input

3
1 2 3

Sample Output

7

题意:找到一个字符串里面所有的不下降子序列的数量之和。

题解:设dp[i] 以a[i]结尾的非递减子串的个数,可以知道 dp[i] = sum(dp[j])+1 (1<=j<i,a[j]<a[i]),这里a[j]<a[i]我们可以直接利用求解逆序数的原理得到,利用树状数组维护整个递推式。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long LL;
const int N = 100005;
const int mod = 1000000007;
int n;
int a[N],b[N],c[N];
int lowbit(int x){
    return x&(-x);
}
void update(int idx,int v){
    for(int i=idx;i<=n;i+=lowbit(i)){
        c[i]=(c[i]+v)%mod;
    }
}
int getsum(int idx){
    int sum = 0;
    for(int i=idx;i>=1;i-=lowbit(i)){
        sum = (sum+c[i])%mod;
    }
    return sum;
}
int main()
{
    while(scanf("%d",&n)!=EOF){
        memset(c,0,sizeof(c));
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            b[i] = a[i];
        }
        sort(b+1,b+1+n);
        int ans = 0;
        for(int i=1;i<=n;i++){
            int idx = lower_bound(b+1,b+1+n,a[i])-b;
            ans=getsum(idx);
            update(idx,ans+1);
        }
        printf("%d
",getsum(n));
    }
    return 0;
}

hdu 2227(树状数组+dp) Find the nondecreasing subsequences

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